Mixing
Inviting friends to dinner (combination/permutation)
$begingroup$
The head of a large company has 9 close friends.
a) In how many ways can he invite six of them to dinner?
b) Repeat a) if two of his friends are divorced, hate each other, and cannot both be invited.
c) Repeat a) if the friends consist of three single people and three married couples and if a husband or wife is invited, the spouse must be invited, too.
Part a is relatively straight forward. It would simply be $C(9,6)$ since the order in which he invites them does not matter.
Part b and c are a little more complicated and I'm not sure how to do them. Any help would be appreciated.
combinatorics permutations
asked Dec 4 '13 at 22:33
micheal walmicheal wal
415
$endgroup$
add a comment |
$begingroup$
The head of a large company has 9 close friends.
a) In how many ways can he invite six of them to dinner?
b) Repeat a) if two of his friends are divorced, hate each other, and cannot both be invited.
c) Repeat a) if the friends consist of three single people and three married couples and if a husband or wife is invited, the spouse must be invited, too.
Part a is relatively straight forward. It would simply be $C(9,6)$ since the order in which he invites them does not matter.
Part b and c are a little more complicated and I'm not sure how to do them. Any help would be appreciated.
combinatorics permutations
asked Dec 4 '13 at 22:33
micheal walmicheal wal
415
$endgroup$
add a comment |
$begingroup$
The head of a large company has 9 close friends.
a) In how many ways can he invite six of them to dinner?
b) Repeat a) if two of his friends are divorced, hate each other, and cannot both be invited.
c) Repeat a) if the friends consist of three single people and three married couples and if a husband or wife is invited, the spouse must be invited, too.
Part a is relatively straight forward. It would simply be $C(9,6)$ since the order in which he invites them does not matter.
Part b and c are a little more complicated and I'm not sure how to do them. Any help would be appreciated.
combinatorics permutations
asked Dec 4 '13 at 22:33
micheal walmicheal wal
415
$endgroup$
The head of a large company has 9 close friends.
a) In how many ways can he invite six of them to dinner?
b) Repeat a) if two of his friends are divorced, hate each other, and cannot both be invited.
c) Repeat a) if the friends consist of three single people and three married couples and if a husband or wife is invited, the spouse must be invited, too.
Part a is relatively straight forward. It would simply be $C(9,6)$ since the order in which he invites them does not matter.
Part b and c are a little more complicated and I'm not sure how to do them. Any help would be appreciated.
combinatorics permutations
combinatorics permutations
asked Dec 4 '13 at 22:33
micheal walmicheal wal
415
asked Dec 4 '13 at 22:33
micheal walmicheal wal
415
asked Dec 4 '13 at 22:33
micheal walmicheal wal
415
asked Dec 4 '13 at 22:33
micheal walmicheal wal
415
asked Dec 4 '13 at 22:33
micheal walmicheal wal
415
415
add a comment |
add a comment |
1 Answer
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$begingroup$
For part b), how many ways can you choose the divorced couple AND four of the other seven individuals. So if the divorced couple is invited, then the remaining four choices are restricted to the remaining seven potential guests. Since we are excluding this case, the number of ways to invite six people is: $C(9,6) - C(7,4)$.
HINT: For part c), you either choose two singletons and 2 couples OR no singletons and 3 couples.
answered Dec 4 '13 at 22:53
Chris KChris K
2,6931814
$endgroup$
$begingroup$
@michaelwal: Please see the hint above. If you're still stuck, I'm willing to help as required.
$endgroup$
– Chris K
Dec 4 '13 at 22:54
$begingroup$
I'm not sure...
$endgroup$
– micheal wal
Dec 4 '13 at 23:14
$begingroup$
@michaelwal: I'll add some details to show you how it's done.
$endgroup$
– Chris K
Dec 4 '13 at 23:15
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
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active
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votes
$begingroup$
For part b), how many ways can you choose the divorced couple AND four of the other seven individuals. So if the divorced couple is invited, then the remaining four choices are restricted to the remaining seven potential guests. Since we are excluding this case, the number of ways to invite six people is: $C(9,6) - C(7,4)$.
HINT: For part c), you either choose two singletons and 2 couples OR no singletons and 3 couples.
answered Dec 4 '13 at 22:53
Chris KChris K
2,6931814
$endgroup$
$begingroup$
@michaelwal: Please see the hint above. If you're still stuck, I'm willing to help as required.
$endgroup$
– Chris K
Dec 4 '13 at 22:54
$begingroup$
I'm not sure...
$endgroup$
– micheal wal
Dec 4 '13 at 23:14
$begingroup$
@michaelwal: I'll add some details to show you how it's done.
$endgroup$
– Chris K
Dec 4 '13 at 23:15
add a comment |
$begingroup$
For part b), how many ways can you choose the divorced couple AND four of the other seven individuals. So if the divorced couple is invited, then the remaining four choices are restricted to the remaining seven potential guests. Since we are excluding this case, the number of ways to invite six people is: $C(9,6) - C(7,4)$.
HINT: For part c), you either choose two singletons and 2 couples OR no singletons and 3 couples.
answered Dec 4 '13 at 22:53
Chris KChris K
2,6931814
$endgroup$
$begingroup$
@michaelwal: Please see the hint above. If you're still stuck, I'm willing to help as required.
$endgroup$
– Chris K
Dec 4 '13 at 22:54
$begingroup$
I'm not sure...
$endgroup$
– micheal wal
Dec 4 '13 at 23:14
$begingroup$
@michaelwal: I'll add some details to show you how it's done.
$endgroup$
– Chris K
Dec 4 '13 at 23:15
add a comment |
$begingroup$
For part b), how many ways can you choose the divorced couple AND four of the other seven individuals. So if the divorced couple is invited, then the remaining four choices are restricted to the remaining seven potential guests. Since we are excluding this case, the number of ways to invite six people is: $C(9,6) - C(7,4)$.
HINT: For part c), you either choose two singletons and 2 couples OR no singletons and 3 couples.
answered Dec 4 '13 at 22:53
Chris KChris K
2,6931814
$endgroup$
For part b), how many ways can you choose the divorced couple AND four of the other seven individuals. So if the divorced couple is invited, then the remaining four choices are restricted to the remaining seven potential guests. Since we are excluding this case, the number of ways to invite six people is: $C(9,6) - C(7,4)$.
HINT: For part c), you either choose two singletons and 2 couples OR no singletons and 3 couples.
answered Dec 4 '13 at 22:53
Chris KChris K
2,6931814
edited Dec 4 '13 at 23:17
answered Dec 4 '13 at 22:53
Chris KChris K
2,6931814
answered Dec 4 '13 at 22:53
Chris KChris K
2,6931814
answered Dec 4 '13 at 22:53
Chris KChris K
2,6931814
2,6931814
$begingroup$
@michaelwal: Please see the hint above. If you're still stuck, I'm willing to help as required.
$endgroup$
– Chris K
Dec 4 '13 at 22:54
$begingroup$
I'm not sure...
$endgroup$
– micheal wal
Dec 4 '13 at 23:14
$begingroup$
@michaelwal: I'll add some details to show you how it's done.
$endgroup$
– Chris K
Dec 4 '13 at 23:15
add a comment |
$begingroup$
@michaelwal: Please see the hint above. If you're still stuck, I'm willing to help as required.
$endgroup$
– Chris K
Dec 4 '13 at 22:54
$begingroup$
I'm not sure...
$endgroup$
– micheal wal
Dec 4 '13 at 23:14
$begingroup$
@michaelwal: I'll add some details to show you how it's done.
$endgroup$
– Chris K
Dec 4 '13 at 23:15
$begingroup$
@michaelwal: Please see the hint above. If you're still stuck, I'm willing to help as required.
$endgroup$
– Chris K
Dec 4 '13 at 22:54
$begingroup$
@michaelwal: Please see the hint above. If you're still stuck, I'm willing to help as required.
$endgroup$
– Chris K
Dec 4 '13 at 22:54
$begingroup$
I'm not sure...
$endgroup$
– micheal wal
Dec 4 '13 at 23:14
$begingroup$
I'm not sure...
$endgroup$
– micheal wal
Dec 4 '13 at 23:14
$begingroup$
@michaelwal: I'll add some details to show you how it's done.
$endgroup$
– Chris K
Dec 4 '13 at 23:15
$begingroup$
@michaelwal: I'll add some details to show you how it's done.
$endgroup$
– Chris K
Dec 4 '13 at 23:15
add a comment |
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